pbuk said:
Thank you
@jbriggs444 for your considered analysis. As it happens I agree with you that
, however that does not mean that option B is a correct answer (and I don't think you are arguing that it is?)
But I do not agree with the rest of your analysis.
I don't agree with this. For me "at rest" at ##t=0## means only that ## t = 0 \implies v = 0 ##.
The mathematical definition of the first derivative is in conflict with this.$$\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$With the understanding that ##a## and ##a+h## are constrained to fall within the domain of ##f##.
If the two one-sided derivatives disagree then the two sided derivative does not exist.
pbuk said:
Where displacement is differentiable that is true, however I do not agree that (in an ideal model) that displacement must be differentiable everywhere: for instance our ideal model of a collision prevents that.
Right. If velocity changes around zero then the position function is not differentiable at zero. So ##v(t)## is undefined at ##t=0##. Not zero. Undefined.
pbuk said:
The mean value theorem requires a differentiable (and therefore continuous) function so you can't invoke it to prove differentiability.
I am invoking it to prove the lack of differentiability.
The mean value theorem requires not just differentiability. It also requires that the function be defined everywhere within a closed interval and that it be differentiable everywhere within the open interval. I am invoking it to prove that the derivative is not defined throughout an open interval that encompasses both positive and negative times. In particular, that it is undefined at ##t=0##.
Let me take you through it. We will be dealing here with horizontal velocity.
I will denote position as ##s(t)## and its first derivative as ##v(t)##.
I will take "at rest" to mean that ##s(t) = 0## throughout some closed interval ending at ##t=0##. This implies that ##v(t)=0## throughout the open interval ending at ##t=0##. I make no inference about ##v(0)##.
Let us denote the start of this interval as ##a##.
I will take "constant velocity" to mean that ##v(t) = k > 0## throughout some open interval beginning at ##t=0##. I make no inference about ##v(0)##.
Let us denote the end of this interval as ##b##.
I think you will grant me that the position function is continuous and well defined over the closed interval ##[a,b]## and that ##s(b) > s(a)##. I think that you will also grant me that the average velocity over the interval is less than ##k##, the constant velocity for ##t > 0##.
That is:$$\frac{s(b) - s(a)}{b-a} < k$$At this point, we have almost everything we need to invoke the mean value theorem. We have a function that is defined and continuous on a closed interval. It is differentiable on the open interval except, possibly at ##t=0##
If ##v(t)## is defined for ##t=0## we will have everything we need for the mean value theorem. So let us suppose (for purposes of a proof by contradiction) that ##v(t)## is defined for ##t=0##.
Then it follows from the mean value theorem that ##v(t) = \frac{s(b) - s(a)}{b-a}## for some ##t## in the open interval ##(a,b)##
But we've already said that ##v(t) = 0## for ##t<0## in this interval. And we've already said that ##v(t) = k## for ##t>0## in this interval.
If ##v(0)= 0## the mean value theorem is falsified.
If ##v(0) = k## the mean value theorem is falsified.
There is a loophole if ##v(0) = \frac{s(b) - s(a)}{b-a}##. This is tedious to address and is obviously silly. So let us ignore it.
Contradiction. So ##v(0)## is undefined.
pbuk said:
It is defined as zero in the question.
No. It is not. The question says "at rest". The implication is that the derivative is zero at ##t < 0##. The implication does not hold for ##t=0##.
pbuk said:
What happens when we cut a string suspending an object?
The acceleration changes suddenly. It is zero before the cut. It is non-zero and constant after the cut. The acceleration at the moment the string is cut is undefined. Two one-sided second derivatives of position exist and are different.
The velocity does not change suddenly. It is zero both before and after the cut. (To be picky, after the cut it is non-zero and increasing with a limit of zero as the time of the cut is approached). The velocity at the time of the cut is well defined and is zero.
The position does not change suddenly.